Explanation:
The given data is as follows.
= 0.483,
= 0.173 M,
= 0.433 M,
= 0.306 M,
= 9.0 atm
According to the ideal gas equation, PV = nRT
or, P =
Also, we know that
Density = 
So, P = MRT
and, M = 
= 
= 
= 0.368 mol/L
Now, we will calculate the cell potential as follows.
E = ![E^{o} - \frac{0.0591}{n} log \frac{[Co^{2+}]^{2}[Cl_{2}]}{[Co^{3+}][Cl^{-}]^{2}}](https://tex.z-dn.net/?f=E%5E%7Bo%7D%20-%20%5Cfrac%7B0.0591%7D%7Bn%7D%20log%20%5Cfrac%7B%5BCo%5E%7B2%2B%7D%5D%5E%7B2%7D%5BCl_%7B2%7D%5D%7D%7B%5BCo%5E%7B3%2B%7D%5D%5BCl%5E%7B-%7D%5D%5E%7B2%7D%7D)
= 
= 
= 
= 0.483 - 0.0185
= 0.4645 V
Thus, we can conclude that the cell potential of given cell at
is 0.4645 V.
actually the answer is C. 15.0 on ed
The cyanide is

A carbon atom has 4 valance electrons and nitrogen has 5. Below is a Lewis-dot-structure of cyanide.
:N≡C.
The carbon atom is still one electron short of having a full octet and so it will seize another electron from almost anything, making the cyanide ion negative and whatever it took the electron from it now positive.
<span>1.
London forces. 4. dipole - dipole. Due to a small hydrogen atom and a
much large fluorine atom, with a large , positive nuclus and large,
negative, p orbitals . This makes it very polar.</span>
Explanation:
It is given that,
Initial orbit of electrons, 
Final orbit of electrons, 
We need to find energy, wavelength and frequency of the wave.
When atom make transition from one orbit to another, the energy of wave is given by :

Putting all the values we get :

We know that : 
So,

Energy of wave in terms of frequency is given by :


Also, 
is wavelength
So,

Hence, this is the required solution.